\newproblem{lay:1_4_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.4.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\mathbf{u}=\begin{pmatrix}7\\2\\5\end{pmatrix}$, $\mathbf{v}=\begin{pmatrix}3\\1\\3\end{pmatrix}$, and $\mathbf{w}=\begin{pmatrix}5\\1\\1\end{pmatrix}$. It
	can be shown that $2\mathbf{u}-3\mathbf{v}-\mathbf{w}=\mathbf{0}$. Use this fact (and no row operations) to find $x_1$ and $x_2$ that satisfy the equation:
	\begin{center}
		$\begin{pmatrix}7 & 3 \\ 2 & 1 \\ 5 & 3\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}5\\1\\1\end{pmatrix}$
	\end{center}
}
{
  % Solution
	We note that the first column of the matrix in the system equation is $\mathbf{u}$, the second column is $\mathbf{v}$ and the vector of independent terms is $\mathbf{w}$.
	Consequently, the equation system is trying to find $x_1$ and $x_2$ such that
	\begin{center}
		$x_1\mathbf{u}+x_2\mathbf{v}=\mathbf{w}$
	\end{center}
	Comparing this equation with the fact of the statement
	\begin{center}
		$2\mathbf{u}-3\mathbf{v}-\mathbf{w}=\mathbf{0} \Rightarrow 2\mathbf{u}-3\mathbf{v}=\mathbf{w}$
	\end{center}
	we deduce that $x_1=2$ and $x_2=-3$.
}
\useproblem{lay:1_4_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
